Convert string into int in C

[screwball]

Hello,
Can you tecll me how to to convert an integer variable into a string variable in C? any help would be greatly appreciated. Thank you .

[aftab1]

» char *itoa(int value, char *buffer, int radix);

Converts an integer value to a null-terminated string using the specified radix and stores the result in the given buffer. If radix is 10 and value is negative the string is preceded by the minus sign (-). With any other radix, value is always considered unsigned.buffer should be large enough to contain any possible value: (sizeof(int)*8+1) for radix=2, i.e. 17 bytes in 16-bits platforms and 33 in 32-bits platforms.

Here is the example:

#include <stdlib.h>	// for itoa() call
#include <stdio.h>	// for printf() call

int main() {
	int num = 123;
	char buf[5];

	// convert 123 to string [buf]
	itoa(num, buf, 10);

	// print our string
	printf("%s\n", buf);

	return 0;
}

[Raiz]

Refer to following thread to get more help:

How to convert string to int

Hope this helps.

[Modifier]

Below is example code for Converting string to int, double and long:

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[])
{
  int i;
  double d;
  long l;

  i = atoi("1");
  l = atol("11111111");
  d = atof("11111.11111");

  printf("%d %ld %f", i, l, d);

  return 0;
}

[gaurav414u]

hey heres a better way ... ( the original way, without using library functions)
you can convert a char type into int type by:
here ch must b a single digit character(0-9) (its not a string)
num=ch-48; // shifting the ASCII value 48 units back.

now use this approach in a loop to convert the whole string as:

int chartoint(char *k)
{
int i;
int l;
int j;
int t=0;
l=strlen(k);

cout<<endl<<l<<endl;
for(i=0;i<l;i++)
{
j=k[i]-48;
t=t+j*(int)pow(10,l-i-1);
}
return t;
}


pass it the string ... say (123) and it will return 123 in integertype;